Problem: Simplify and expand the following expression: $ \dfrac{4}{q + 8}- \dfrac{1}{q - 4}- \dfrac{5q}{q^2 + 4q - 32} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{5q}{q^2 + 4q - 32} = \dfrac{5q}{(q + 8)(q - 4)}$ Now we have: $ \dfrac{4}{q + 8}- \dfrac{1}{q - 4}- \dfrac{5q}{(q + 8)(q - 4)} $ The least common multiple of the denominators is: $ (q + 8)(q - 4)$ In order to get the first term over $(q + 8)(q - 4)$ , multiply by $\dfrac{q - 4}{q - 4}$ $ \dfrac{4}{q + 8} \times \dfrac{q - 4}{q - 4} = \dfrac{4(q - 4)}{(q + 8)(q - 4)} $ In order to get the second term over $(q + 8)(q - 4)$ , multiply by $\dfrac{q + 8}{q + 8}$ $ \dfrac{1}{q - 4} \times \dfrac{q + 8}{q + 8} = \dfrac{q + 8}{(q + 8)(q - 4)} $ Now we have: $ \dfrac{4(q - 4)}{(q + 8)(q - 4)} - \dfrac{q + 8}{(q + 8)(q - 4)} - \dfrac{5q}{(q + 8)(q - 4)} $ $ = \dfrac{ 4(q - 4) - (q + 8) - 5q} {(q + 8)(q - 4)} $ Expand: $ = \dfrac{4q - 16 - q - 8 - 5q}{q^2 + 4q - 32} $ $ = \dfrac{-2q - 24}{q^2 + 4q - 32}$